$$ \dfrac{ab^2}{c^2} + \dfrac{bc^2}{a^2} + \dfrac{ca^2}{b^2} + a + b + c \ge \dfrac{6(a^2 + b^2 + c^2)}{a + b + c} \,\,\, (1)$$
Giải
Bất đẳng thức ban đầu tương đương:$(a + b + c)(\dfrac{ab^2}{c^2} + \dfrac{bc^2}{a^2} + \dfrac{ca^2}{b^2}) + (a + b + c)^2 \geq 6(a^2 + b^2 + c^2)$
Trước tiên, ta cần chứng minh:
$\dfrac{ab^2}{c^2} + \dfrac{bc^2}{a^2} + \dfrac{ca^2}{b^2} \geq a + b + c$
Thật vậy, ta có:
$\dfrac{ab^2}{c^2} + \dfrac{ab^2}{c^2} + \dfrac{bc^2}{a^2} + c + c \geq 5 \sqrt[5]{\dfrac{a^2b^5c^4}{a^2c^4}} = 5b$
Tương tự:
$\dfrac{bc^2}{a^2} + \dfrac{bc^2}{a^2} + \dfrac{ca^2}{b^2} + 2a \geq 5c$
$\dfrac{ca^2}{b^2} + \dfrac{ca^2}{b^2} + \dfrac{ab^2}{c^2} + 2b \geq 5a$
$\Rightarrow 3(\dfrac{ab^2}{c^2} + \dfrac{bc^2}{a^2} + \dfrac{ca^2}{b^2}) \geq 3(a + b + c)$
$\Leftrightarrow \dfrac{ab^2}{c^2} + \dfrac{bc^2}{a^2} + \dfrac{ca^2}{b^2} \geq a + b + c$
Do đó:
$VT_{(1)} \geq 2(a + b + c)^2 \geq 2.3(a^2 + b^2 + c^2) = 6(a^2 + b^2 + c^2) = VF$
Dấu "=" xảy ra khi: a = b = c